Ajuda,obrigado.
2x²/(x² - 4) > 2/(x + 2) → you can simplify by 2 both sides
x²/(x² - 4) > 1/(x + 2) → do not use the cross-multiply, it's a trap
[x²/(x² - 4)] - [1/(x + 2)] > 0 → you know that: x² - 4 = x² - 2² = (x + 2).(x - 2)
[x²/(x + 2).(x - 2)] - [1/(x + 2)] > 0
[x²/(x + 2).(x - 2)] - [(x - 2)/(x + 2).(x - 2)] > 0
[x² - (x - 2)]/[(x + 2).(x - 2)] > 0 → where: x ≠ - 2 and where: x ≠ 2
(x² - x + 2)/[(x + 2).(x - 2)] > 0
Even if this polynomial is a fraction, its sign will be the same that the product, i.e.:
(x² - x + 2).(x + 2).(x - 2) > 0 ← memorize this result
= x² - x + 2
= x² - x + [(1/2)² - (1/2)²] + 2
= x² - x + (1/2)² - (1/2)² + 2
= x² - x + (1/2)² - (1/4) + (8/4)
= [x² - x + (1/2)²] + (7/4)
= [x - (1/2)]² + (7/4)
A square is always positive or zero, but if you add to it (7/4), the result will be always positive.
x² - x + 2 > 0
Restart from the memorized result
(x² - x + 2).(x + 2).(x - 2) > 0 → we've just seen that: (x² - x + 2) > 0
(x + 2).(x - 2) > 0 → the roots are: - 2 ; 2 → then you make a table
x_____-∞____-2____2____+∞
(x + 2)____-__0__+____+
(x - 2)____-_____-__0__+
sign______+__0_-__0__+
…and you can see when the sign is > 0
x Є ] -∞ ; - 2 [ U ] 2 ; +∞ [
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Answers & Comments
2x²/(x² - 4) > 2/(x + 2) → you can simplify by 2 both sides
x²/(x² - 4) > 1/(x + 2) → do not use the cross-multiply, it's a trap
[x²/(x² - 4)] - [1/(x + 2)] > 0 → you know that: x² - 4 = x² - 2² = (x + 2).(x - 2)
[x²/(x + 2).(x - 2)] - [1/(x + 2)] > 0
[x²/(x + 2).(x - 2)] - [(x - 2)/(x + 2).(x - 2)] > 0
[x² - (x - 2)]/[(x + 2).(x - 2)] > 0 → where: x ≠ - 2 and where: x ≠ 2
(x² - x + 2)/[(x + 2).(x - 2)] > 0
Even if this polynomial is a fraction, its sign will be the same that the product, i.e.:
(x² - x + 2).(x + 2).(x - 2) > 0 ← memorize this result
= x² - x + 2
= x² - x + [(1/2)² - (1/2)²] + 2
= x² - x + (1/2)² - (1/2)² + 2
= x² - x + (1/2)² - (1/4) + (8/4)
= [x² - x + (1/2)²] + (7/4)
= [x - (1/2)]² + (7/4)
A square is always positive or zero, but if you add to it (7/4), the result will be always positive.
x² - x + 2 > 0
Restart from the memorized result
(x² - x + 2).(x + 2).(x - 2) > 0 → we've just seen that: (x² - x + 2) > 0
(x + 2).(x - 2) > 0 → the roots are: - 2 ; 2 → then you make a table
x_____-∞____-2____2____+∞
(x + 2)____-__0__+____+
(x - 2)____-_____-__0__+
sign______+__0_-__0__+
…and you can see when the sign is > 0
x Є ] -∞ ; - 2 [ U ] 2 ; +∞ [